How do you write the partial fraction decomposition of the rational expression  (x^4+6)/(x^5+7x^3)?

Nov 14, 2016

The answer is $= \frac{- \frac{6}{49}}{x} + \frac{\frac{6}{7}}{x} ^ 3 + \frac{\frac{55}{49} x}{{x}^{2} + 7}$

Explanation:

$\frac{{x}^{4} + 6}{{x}^{5} + 7 {x}^{3}} = \frac{{x}^{4} + 6}{\left({x}^{3}\right) \left({x}^{2} + 7\right)}$

Let's do the decompositiom in partial fractions

$\frac{{x}^{4} + 6}{\left({x}^{3}\right) \left({x}^{2} + 7\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x} ^ 3 + \frac{D x + E}{{x}^{2} + 7}$

$= \frac{A {x}^{2} \left({x}^{2} + 7\right) + B x \left({x}^{2} + 7\right) + C \left({x}^{2} + 7\right) + {x}^{3} \left(D x + E\right)}{\left({x}^{3}\right) \left({x}^{2} + 7\right)}$

Therefore,

${x}^{4} + 6 = A {x}^{2} \left({x}^{2} + 7\right) + B x \left({x}^{2} + 7\right) + C \left({x}^{2} + 7\right) + {x}^{3} \left(D x + E\right)$

Let $x = 0$$\implies$ $6 = 7 C$ $\implies$ $C = \frac{6}{7}$

Coefficients of ${x}^{2}$, $\implies$ $0 = 7 A + C$ ; $A = - \frac{6}{49}$

coefficients of ${x}^{4}$ ; $\implies$ ; $1 = A + D$ ; $D = 1 + \frac{6}{49} = \frac{55}{49}$

Coefficients of ${x}^{3}$ ; $\implies$ ; $0 = B + E$

Coefficients of $x$ ; $\implies$ ; $0 = 7 B$ $\implies$ $B = 0$

$E = 0$

$\frac{{x}^{4} + 6}{\left({x}^{3}\right) \left({x}^{2} + 7\right)} = \frac{- \frac{6}{49}}{x} + \frac{0}{x} ^ 2 + \frac{\frac{6}{7}}{x} ^ 3 + \frac{\frac{55}{49} x + 0}{{x}^{2} + 7}$